Question

In a certain town, $60\%$ of the families own a car, $30\%$own a house and $20\%$ own both a car and a house. If a family is randomly chosen, what is the probability that this family owns a car or a house but not both ?

• A. 0.5
• B. 0.7
• C. 0.1
• D. 0.9

Answer 1

Answer: (A)

0.5

Answer 2

Let $A$ be the set of families who own a car and $B$
be the set of families who own a house.
Then, $P(A)=60 \%, P(B)=30 \%$
and $P(A \cap B)=20 \%$
Now, $P(A \cup B) =P(A)+P(B)-P(A \cap B)$
$=60+30-20=70 \%$
Now, families who owns a car or a house but not both $=A \Delta B$
$\Rightarrow P(A \Delta B) =P(A \cup B)-P(A \cap B)$
$=70-20=50 \%=0.5$